Menu Close

solution-Q1-a-n-ln-m-m-0-ln0-5-ln-12-75-ln0-5-2-64-N-N-0-0-5-n-6-02-10-23-0-5-2-64-9-66-10-22-A-N-1-5-10-4-9-66-10-22-1-45-10-19-Bq-

Tinku Tara 0 Comments
Pin




Question Number 94756 by AshrafNejem last updated on 20/May/20
solution: Q1)a) n=((ln(m/m_0 ))/(ln0.5)) = ((ln(12/75))/(ln0.5)) =2.64 N=N_0 (0.5)^n = 6.02×10^(23) (0.5)^(2.64) = 9.66×10^(22) A=λN=1.5×10^(−4) × 9.66×10^(22) =1.45×10^(19) Bq
$$\left.{s}\left.{olution}:\:\mathrm{Q1}\right){a}\right)\:{n}=\frac{{ln}\left({m}/{m}_{\mathrm{0}} \right)}{{ln}\mathrm{0}.\mathrm{5}}\:=\:\frac{{ln}\left(\mathrm{12}/\mathrm{75}\right)}{{ln}\mathrm{0}.\mathrm{5}}\:=\mathrm{2}.\mathrm{64} \\ $$$${N}=\mathrm{N}_{\mathrm{0}} \left(\mathrm{0}.\mathrm{5}\right)^{{n}} \:=\:\mathrm{6}.\mathrm{02}×\mathrm{10}^{\mathrm{23}} \left(\mathrm{0}.\mathrm{5}\right)^{\mathrm{2}.\mathrm{64}} =\:\mathrm{9}.\mathrm{66}×\mathrm{10}^{\mathrm{22}} \\ $$$${A}=\lambda{N}=\mathrm{1}.\mathrm{5}×\mathrm{10}^{−\mathrm{4}} \:×\:\mathrm{9}.\mathrm{66}×\mathrm{10}^{\mathrm{22}} =\mathrm{1}.\mathrm{45}×\mathrm{10}^{\mathrm{19}} \:{Bq} \\ $$
Commented by mr W last updated on 20/May/20
maybe only you know for which question your answer is. therefore please post your answer in the same thread as the question.
$${maybe}\:{only}\:{you}\:{know}\:{for}\:{which} \\ $$$${question}\:{your}\:{answer}\:{is}.\:{therefore} \\ $$$${please}\:{post}\:{your}\:{answer}\:{in}\:{the}\:{same} \\ $$$${thread}\:{as}\:{the}\:{question}. \\ $$
Commented by MJS last updated on 20/May/20
o tempora o mores
$$\mathrm{o}\:\mathrm{tempora}\:\mathrm{o}\:\mathrm{mores} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *